Download CHAPTER 4 RANDOM VARIABLES AND DISTRIBUTIONS

CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
Every baby is born either male or female. So every
child is the outcome of a human binomial experiment.
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HS_PS_S1_04_CO.indd 130
8/31/11 6:49 PM
In a family with one child, what is the chance that the child is a
girl? In a family with two children, what is the chance that both
of them are girls? What about the chance of having three girls,
or four, or five? Random variables can help describe situations
like these.
In This Chapter
You will learn how random variables and probability distributions can describe
problem situations. You’ll also learn how to use expected value to make decisions.
Specific discrete and continuous distributions, such as binomial and normal
distributions, will be covered as well as techniques for using the standard normal
distribution to compare outcomes.
Topic List
► Chapter 4 Introduction
► Creating Probability Distributions
► Interpreting Probability Distributions
► Expected Value
► Binomial Distributions
► Continuous Random Variables
► The Normal Distribution
► Standardizing Data
► Comparing Scores
► The Standard Normal Curve
► Finding Standard Scores
► Chapter 4 Wrap-Up
RANDOM VARIABLES AND DISTRIBUTIONS
131
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HS_PS_S1_04_CO.indd 131
8/31/11 6:49 PM
Chapter 4 Introduction
Though individual events can be random, they can
yield patterns when repeated multiple times.
Probability Distributions
The relative frequency histogram shows the proportions for the number of
boys in a family with 3 children, given the predicted probability of 50%.
According to this graph, there is a 37.5% chance of having exactly 1 boy out
of 3 children. You can also see that there is a 25% chance that all 3 children
will have the same gender (just add the heights of the first and last bar).
40
35
30
Probability (%)
It iiss of
oft
often
ten predicted that the probability that a newborn baby will be a
boy (or girl, as the case may be) is 50%. Due to the natural gender ratio,
however, the actual probability that a newborn will be a boy is not exactly
50%.
% In the United States,, for example,
p , the chances are slightly
g y higher
g
that
a newborn will be a boy rather than a girl.
Number of Boys in Family
(Predicted)
25
20
15
10
5
0
The graph for the actual number of boys in a family with 3 children,
however, tells a slightly different story. The relative frequency histogram
estimates the actual proportions considering the natural gender ratio.
0
1
2
Number of boys
3
Number of Boys in Family
(Actual)
Notice that, compared to the first graph, the chances are greater that a
family of 3 will have mostly boys (as indicated by the higher bars for 2 and
3 boys). Although there are slightly more all-boy families than would be
predicted by pure probability, the histograms are quite similar.
40
35
Probability (%)
30
25
20
15
10
5
0
0
132
CHAPTER 4
1
2
Number of boys
3
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_IN_RG.indd 132
9/17/11 2:24 AM
Chapter 4 Introduction
The
Th
he Normal Distribution
When
W
Wh
hen a ddata set has a normal distribution, a histogram that represents the
data
at set is symmetric and has a bell shape.
The Normal Distribution
The
h symmetry of this curve demonstrates that more data are near the mean
th
tha
an iin th
the outlying
tl i regions
i
than
regions.
The normal distribution has properties that are very useful in statistics.
When data are normally distributed, about 68% of the data values are
within a range of 1 standard deviation above and below the mean.
Mean
For example, suppose the average length of a full-term newborn baby
is 20.1 inches with a standard deviation of 0.4 inches. Because lengths
of babies are normally distributed, we can assume that roughly 68% of
newborn babies are between 19.7 and 20.5 inches in length. These values
are obtained by subtracting and adding 0.4 from 20.1.
68% of newborn
babies
19.3
19.7
20.1
20.5
20.9
Other properties of the normal distribution will be explored later in the chapter.
Applying It
In this chapter, you will learn how to create and
interpret probability and binomial distributions
using both tables and histograms. You will work
with real data to learn about expected value and
continuous random variables. You will also learn
about normal distributions and how the standard
normal curve is used to interpret data and make
predictions.
CHAPTER 4 INTRODUCTION
133
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HS_PS_S1_04_IN_RG.indd 133
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Preparing for the Chapter
Review the following skills to prepare for the concepts
in Chapter 4.
►
Find probabilities of events.
►
Interpret histograms.
►
Determine areas of figures in the coordinate plane.
x−a
► Solve linear equations in the form c = _____ .
b
► Simplify the expression 1 − x for different values of x.
Problem Set
A stack of 22 cards are numbered 1 to 22. If a card is randomly
selected, find the probability of the event.
1.
The number 4 is selected.
2.
An even number is selected.
3.
An odd number less than 8 is selected.
4.
A prime number or an even number is selected.
5.
A number less than 11 is selected.
6.
A number less than 17 or greater than 20 is selected.
The histogram shows the number of children in 32 selected families.
How many families have 5 children?
8.
How many families have 2 or more children?
9.
How many families have from 1 to 3 children?
10.
What percent of the families have from 2 to
4 children?
Number of Children per Family
12
10
Families
7.
8
6
4
2
0
1
134
CHAPTER 4
2
3
4
Children
5
6
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_IN_PS.indd 134
9/24/11 1:54 AM
Chapter 4 Introduction
Determine the area of the shaded figure.
11.
13.
x
6
7
6
5
4
3
2
1
5
4
3
2
1
y
0
y
0
12.
x
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7
14.
x
6
5
4
3
2
1
0
x
5
4
3
2
1
y
1 2 3 4 5 6 7 8 9
0
y
1 2 3 4 5 6 7 8 9 10 11
Determine the value of x.
15.
x − 30
1.1 = ______
3
17.
x − 120
−1.15 = _______
4.8
16.
x − 1.22
2.05 = ________
0.1
18.
x − 450
−2.5 = _______
12
Find the value of the expression 1 − x for the given value of x.
19.
3
x = __
8
21.
2
x = __
9
23.
x = 0.381
20.
x = 0.92
22.
x = 0.3072
24.
x = 0.0011
CHAPTER 4 INTRODUCTION
135
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HS_PS_S1_04_IN_PS.indd 135
9/17/11 2:29 AM
Creating Probability
Distributions
Probabilities of all outcomes of a discrete random
variable can be summarized in a table.
Creating Probability Distribution Tables
A study of a new treatment for high cholesterol undergoes clinical testing
with the results summarized in this frequency table.
X
0–9
10–19
20–29
30–39
40–49
f
81
73
35
47
14
In this table, X stands for the number of points that the patient’s cholesterol level
decreased, and f stands for the number of patients. For instance, 73 patients had
their cholesterol decrease between 10 and 19 points after the new treatment.
A patient wants to know the probability that the new treatment will lower
his cholesterol from 20 to 29 points. Here, a probability distribution table
would be useful because it shows the probabilities for all of the outcomes.
REMEMBER
The letter X is used to represent
a discrete random variable. For
example, when a coin is tossed,
the variable X would represent
the two outcomes: heads and
tails. These outcomes are both
discrete and random.
HOW TO CREATE A PROBABILITY DISTRIBUTION TABLE
To create a probability distribution table from a frequency table
Step 1 Find the sum of the frequencies.
Step 2 Divide each frequency by the sum found in Step 1.
Step 3 Replace each frequency in the table with the ratio found in Step 2.
Step 4 Change the label f to P.
For the frequency table describing the clinical testing, the total number of
frequencies is 81 + 73 + 35 + 47 + 14 = 250. So 250 patients took part in
the study. Here is the complete probability distribution table.
X
0–9
10–19
20–29
30–39
40–49
THINK ABOUT IT
P
0.324
0.292
0.14
0.188
0.056
The sum of the probabilities in
a probability distribution table
always equals one.
According to this table, the probability that the new treatment will lower a
patient’s cholesterol between 20 and 29 points is 0.14 or 14%.
136
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T01_RG.indd 136
9/17/11 2:33 AM
In a different study, patients are separated into four groups according
to a spinner.
4
For the spinner, there are four different outcomes:
Group 1, Group 2, Group 3, and Group 4. There are
6 sections overall.
1
2
Notice that some outcomes are more likely than
others. For instance, there are two sections for groups
1 and 2, but only one section for groups 3 and 4.
2
1
3
The probability distribution table shows that the chances of being in Group 1
or Group 2 are higher than the chances of being in Group 3 or Group 4.
X
P
1
2
3
4
1
__
1
__
1
__
1
__
3
3
6
6
Properties of Probability Distributions
There are two properties that every probability distribution table has.
PROPERTIES OF PROBABILITY DISTRIBUTIONS
Every probability distribution for a discrete random variable has the
following properties:
• Each probability is a number between 0 and 1.
• The sum of all probabilities is equal to 1.
This probability distribution table violates the first property because each
probability is not a number between 0 and 1. So it is not a probability
distribution.
X
1
2
3
4
5
P
0.5
0.1
0.3
0.4
−0.3
THINK ABOUT IT
Every probability distribution for
a discrete random variable must
satisfy both properties.
This probability distribution table violates the second property because
the sum of all probabilities is not equal to 1. So it is not a probability
distribution.
X
5
10
15
20
25
30
P
0.34
0.5
0.101
0.32
0.34
0.122
CREATING PROBABILITY DISTRIBUTIONS
137
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HS_PS_S1_04_T01_RG.indd 137
9/24/11 1:57 AM
Problem Set
A survey asked a group of people the size of their households.
The results are in the table.
Household Size
1
2
3
4
5
ƒ
4
12
6
2
1
1.
How many people were surveyed?
2.
How many people reported a household size of 4?
3.
Complete the probability distribution table.
1
Household Size
2
3
4
5
0.08
P
Match the frequency table with the probability distribution table.
A.
4.
138
X
P
0
B.
X
P
0.1
0
1
0.12
2
C.
X
P
0.1
0
1
0.2
0.32
2
3
0.16
4
0.3
X
ƒ
0
D.
X
P
0.2
0
0.2
1
0.24
1
0.2
0.3
2
0.1
2
0.2
3
0.1
3
0.26
3
0.2
4
0.3
4
0.2
4
0.2
X
ƒ
X
ƒ
X
ƒ
5
0
10
0
10
0
5
1
10
1
12
1
10
1
6
2
15
2
5
2
10
2
16
3
5
3
13
3
10
3
8
4
15
4
10
4
10
4
15
CHAPTER 4
5.
6.
7.
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T01_PS.indd 138
9/17/11 2:45 AM
Create a probability distribution table.
8.
6
10.
In a medical study, the heights of a group of
men were measured to the nearest inch. There
were 120 men with heights of 60–64 inches,
280 men with heights of 65–69 inches, 312 men
with heights of 70–74 inches, and 162 men with
heights of 75–79 inches.
11.
The Mosquitoes are playing the Gnats in a
three-game series. The probability is 0.25 that
the Gnats will win all 3 games, 0.4 that they will
win 2 of the games, 0.2 that they will win 1 of
the games, and 0.15 that they will not win any
of the games.
12.
What is the probability of rolling a 1, 2, 3, 4, 5,
or 6 when rolling a 6-sided number cube?
13.
There is a 40% chance it will rain tomorrow, a
30% chance it will snow, and a 30% chance it
will be partly cloudy.
5
4
3
2
1
0
Yellow
9.
Red
Orange
Blue
Green
10
9
8
7
6
5
4
3
2
1
0
Mon
Tue
Wed
Thu
Fri
Sat
Sun
Explain why the table is not a probability distribution.
14.
Result
P
0
15.
Length
P
0.25
5–14
0.25
1
0.45
15–24
-0.3
2
0.0
25–34
0.4
3
0.35
35–44
0.5
45–54
0.15
Find the value of n using the probability distribution table.
16.
Color
P
red
blue
1
__
1
__
3
8
gold
silver
n
1
__
6
17.
Number
of Guests
per Room
0
1
2
3
4
P
0.42
0.31
0.16
n
0.09
Solve.
18.
Challenge Create a probability distribution for rolling two 6-sided number cubes,
where X is the sum of the numbers on the number cube, and P is the probability of
obtaining the sum.
CREATING PROBABILITY DISTRIBUTIONS
139
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HS_PS_S1_04_T01_PS.indd 139
9/24/11 1:58 AM
Interpreting Probability
Distributions
The likelihood of many events can be summarized
by a table or histogram.
Interpreting Probability Distribution Tables
A summer camp program director is planning activities for the summer.
The probability distribution table shows the probability P that the age of a
randomly chosen camper is X.
X
10
11
12
13
14
P
0.15
0.3
0.2
0.25
0.1
For the first activity, the program director selects a camper at random to
read announcements. What is P (X = 13): the probability that the camper
selected is 13 years old?
According to the probability distribution table, when X = 13, the probability
is P = 0.25. So, P(X = 13) = 0.25.
In other words, the probability of selecting a 13-year-old camper at random
is 0.25, or 25%.
Suppose the program director wants to determine P(X ≥ 13), the probability
that a randomly selected camper is at least 13 years old.
This can be answered by adding individual probabilities.
P (X ≥ 13) = P (X = 13) + P (X = 14)
= 0.25 + 0.1
= 0.35
There is a 35% probability of selecting a camper who is at least 13 years old.
140
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T02_RG.indd 140
9/24/11 2:00 AM
Interpreting Probability Distribution Histograms
The program director now has to decide which staff members to assign to
which activities. For this, she is using this probability distribution histogram.
P
Ages of Staff Members
0.5
REMEMBER
0.4
A histogram is a graph showing
the frequency distribution
of data.
0.3
0.2
0.1
0
6
12
18
24
30
36
42
48
54
X
In this histogram, X stands for the ages of individual staff members.
P is the probability that if a single staff member is chosen at random,
then the staff member will be that age.
For the first activity, the program director selects a staff member at random
to lead the songs at dinner. What is P (18 ≤ X < 24): the probability that the
staff member selected is at least 18 years old and less than 24 years old?
The first bar of the histogram represents 18 ≤ X < 24. The height of this
bar is at the value P = 0.2. So the probability of selecting a staff member
who is at least 18 years old and less than 24 years old is 0.2%, or 20%.
Probability distribution histograms can help determine more than just
single entries. Suppose the program director wants to determine P (X ≥ 30),
the probability that a randomly selected staff member is at least 30 years old.
This can be answered by adding individual probabilities.
P (X ≥ 30) = P (30 ≤ X < 36) + P (36 ≤ X < 42)
= 0.3 + 0.1
= 0.4
There is a 40% probability of selecting a staff member who is at least
30 years old.
INTERPRETING PROBABILITY DISTRIBUTIONS
141
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HS_PS_S1_04_T02_RG.indd 141
9/24/11 2:00 AM
Problem Set
Use the probability distribution table to find the probability of the event.
X
0
1
2
3
4
5
P
0.3
0.05
0.1
0.15
0.15
0.25
1.
P (X = 3)
5.
P (X > 2 or X = 0)
2.
P (X = 2 or X = 0)
6.
P (X ≠ 3)
3.
P (1 < X ≤ 3)
7.
P (X < 4)
4.
P (0 < X < 3)
8.
P (X ≥ 4)
Use the probability histogram to find the probability of the event.
0.25
0.2
P
0.15
0.1
0.05
0
1
2
3
4
X
5
6
7
P (X = 6)
13.
P (X = 1 or X > 3)
10.
P (X = 1 or X = 3)
14.
P (X ≠ 6)
11.
P (1 < X ≤ 6)
15.
P (X > 4)
12.
P (0 < X < 4)
16.
P (X ≤ 5)
9.
The probability distribution histogram shows the weight distribution of
hamadryas baboons that live in a certain region. Find the probability of
the event.
Weights of Hamadryas Baboons
0.3
0.25
P
0.2
0.15
0.1
0.05
5
10 15
20 25 30
Weight (lb)
35
40
45
17.
P (5 ≤ X < 20)
20.
P (X < 30)
18.
P (25 ≤ X < 35)
21.
P (X < 15)
19.
P (X < 15 or X ≥ 25)
22.
P (X < 20 or X ≥ 30)
142
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T02_PS.indd 142
9/17/11 2:58 AM
Each week, Evan randomly picks a weekday (Monday through Friday)
to visit the library.
23.
Create a probability distribution table for this
situation. What is the probability that he does
not visit the library on Friday?
24.
Let X be the number of times he visits the library
on Friday for the next two weeks. Create a
probability distribution table for X. Find P (X > 0).
A message center receives calls from 6 p.m. until 6 a.m. Typically, 50%
of the calls come in between 6 p.m. and 9 p.m.; 25% of the calls come
in between 9 p.m. and midnight; 20% of the calls come in between
midnight and 3 a.m.; and 5% of the calls come in between 3 a.m.
and 6 a.m.
25.
What is the probability that a call will come in
between 6 p.m. and 9 p.m.?
26.
What is the probability that a call will come in
between 6 p.m. and midnight?
27.
What is the probability that a call will come in
between midnight and 6 a.m.?
28.
What is the probability that a call will come in
between 9 p.m. and 6 a.m.?
Use the Weights of Hamadryas Baboons probability histogram to find
the probability of the event.
29.
Challenge Two hamadryas baboons from the region are
randomly selected. What is the probability that both weigh less
than 30 pounds?
INTERPRETING PROBABILITY DISTRIBUTIONS
143
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HS_PS_S1_04_T02_PS.indd 143
9/17/11 2:58 AM
Expected Value
Probability distributions can tell us what to expect
when all outcomes are considered.
Using Probability for Estimation
Each week the manager at a large restaurant orders eggs that are used
for breakfast items at the restaurant. The probability distribution table
shows the probability P that a customer will order a breakfast item with
X number of eggs.
The food manager expects about 800 customers
to have breakfast each week at the restaurant.
How many of these customers are expected to
order breakfast items with two eggs?
According to the probability distribution table,
the probability that a customer will order a
breakfast with two eggs is 0.42, or 42%.
X
P
0
0.29
1
0.18
2
0.42
3
0.11
REMEMBER
In a probability distribution
table, the sum of all probabilities
is equal to 1, and each individual
probability is a number between
0 and 1.
Out of 800 patrons, 42% are expected to order
breakfast items with two eggs. So about 0.42 · 800 = 336 customers
are expected to order breakfast items with two eggs.
Finding Expected Value
The prediction above is based on a single outcome—that a customer will
order a breakfast item that has two eggs.
What if the restaurant manager wants to know the average number of eggs
for all breakfast items served at the restaurant?
The manager needs to know the expected value—the expected number of
eggs in a typical breakfast served at the restaurant.
EXPECTED VALUE
Suppose each outcome of a random variable X is x1, x2, . . . ., xn and has
probability p1, p2, . . . . , pn, respectively. The average, or expected value,
of X is
E(X ) = x1 · p1 + x2 · p2 + . . . + xn · pn.
144
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T03_RG.indd 144
9/24/11 2:05 AM
The expected value is the sum of the products of the individual
probabilities and their respective probabilities.
E(X ) = x1 · p1 + x2 · p2 + x3 · p3 + x4 · p4
= 0 · (0.29) + 1 · (0.18) + 2 · (0.42) + 3 · (0.11)
= 0 + 0.18 + 0.84 + 0.33
= 1.35
The expected value of 1.35 means that the food manager can expect that
each breakfast ordered will have 1.35 eggs on average.
THINK ABOUT IT
With an expected value of
1.35 eggs per breakfast, the
manager can predict that about
1.35 · (800) = 1080 eggs will be
consumed in a week.
Solving Problems with Expected Value
The recommended amount of riboflavin per day is 1.7 milligrams. Suppose
there is a 30% chance that a person will take 0 milligrams of riboflavin
per day; a 40% chance that a person will take 1 milligram of riboflavin per
day; a 20% chance that a person will take 2 milligrams of riboflavin per
day; a 5% chance that a person will take 3 milligrams of riboflavin per day;
and a 5% chance that a person will take 4 milligrams of riboflavin per day.
What is the expected value of riboflavin intake per day?
BY THE WAY
Riboflavin is also known as
vitamin B2.
E(X ) = x1 · p1 + x2 · p2 + x3 · p3 + x4 · p4
= 0 · (0.3) + 1 · (0.4) + 2 · (0.2) + 3 · (0.05) + 4 · (0.05)
= 0 + 0.4 + 0.4 + 0.15 + 0.2
= 1.15
So the expected value of riboflavin intake for a single person is about
1.15 milligrams per day. This value is below the recommended amount
of 1.7 milligrams per day.
EXPECTED VALUE
145
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HS_PS_S1_04_T03_RG.indd 145
9/24/11 2:05 AM
Problem Set
The table shows the distribution of household sizes in a certain city.
1.
2.
3.
If there were 6000 households in the city,
how many households would you expect to be
of size 1?
Household Size
P
1
0.16
If there were 4000 households in the city,
how many households would you expect to be
of size 4?
2
0.48
3
0.24
What is the expected value of household size in
this city?
4
0.08
5
0.04
Determine the expected value of X.
4.
5.
146
X
P
6.
X
P
0
8.
X
P
0.1
1998
0.42
1
0.1
1999
0.2
0
1
__
1
1
__
3
2
0.1
2000
0.15
2
1
__
3
0.1
2001
0.1
1
__
4
0.1
2002
0.13
3
5
0.5
X
P
X
P
0
0.5
1998
0.2
1
0.1
2000
0.2
X
3
6
6
P
7.
9.
0
1
__
6
1
__
3
2
0.1
2002
0.2
12
1
__
3
0.1
2004
0.2
4
0.1
2006
0.2
18
1
__
5
0.1
CHAPTER 4
3
6
6
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T03_PS.indd 146
9/17/11 3:47 AM
Solve.
10.
A clothing store has a section of jeans on sale.
There is a 30% chance of picking a pair with a
30-inch inseam; a 25% chance of picking a pair
with a 32-inch inseam; a 10% chance of picking
a pair with a 34-inch inseam; a 10% chance of
picking a pair with a 36-inch inseam; a 10% chance
of picking a pair with a 38-inch inseam; and a
15% chance of picking a pair with a 40-inch inseam.
Create a probability distribution table and find the
expected value of the inseam length of the jeans.
11.
Raina is entering a 64-player cribbage
tournament. The probability of her playing
1
1 ; exactly 2 matches is __
exactly 1 match is __
2
4;
1 ; exactly 4 matches is ___
1
exactly 3 matches is __
8
16 ;
1
exactly 5 matches is ___
32 ; and exactly 6 matches
1
is ___
32 . Create a probability distribution table for
this situation, and find the expected value of the
number of matches Raina will play.
The Columbus Crew of Major League Soccer had the following results last year.
Number of Goals
P
0
1
2
3
1
__
1
__
3
___
1
__
3
5
10
12.
What is the expected value of the number of
goals the Columbus Crew will score in a match?
13.
If the Columbus Crew played 30 matches,
convert the probability distribution table into a
frequency table.
6
14.
Determine the mean number of goals the
Columbus Crew scored in a match. Explain why
this is greater than, less than, or equal to the
expected value.
Use the probability distribution table.
X
P
6
m
8
1
__
n
1
__
15.
2
4
Challenge If the expected value of this distribution is 8.5, find the
value of m and n.
EXPECTED VALUE
147
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HS_PS_S1_04_T03_PS.indd 147
9/17/11 3:47 AM
Binomial Distributions
A table or graph can be used to represent outcomes
of a variable that has a binomial distribution.
Determining Whether a Variable Has a Binomial
Distribution
A doctor knows that for a certain type of adult migraine headache,
Medicine Q works about 70% of the time. If the medicine is tried with two
different patients with this condition, the resulting probability distribution is
a binomial distribution.
BINOMIAL DISTRIBUTION
A variable has a binomial distribution if
• There are a fixed number of independent trials.
• The outcome of each trial is success or failure.
• The probability of success for each trial is the same.
This situation meets all of the conditions for a binomial distribution.
• There are a fixed number of independent trials; there are two trials, and
each patient taking Medicine Q is independent of the other.
• The outcome of each trail is a success or failure; the medicine will help
the migraine (success) or it will not (failure).
THINK ABOUT IT
Two or more coin tosses have
a binomial distribution. The
probability of success and failure
is the same—50%.
• The probability of success for each trial is the same; there is a 0.7
probability of success when a patient takes Medicine Q.
Creating Binomial Distributions
Suppose later in the day, the doctor is seeing two patients who have this
type of adult migraine headache. We can create a binomial distribution table
or graph to represent outcomes for these two patients.
The variable X can take on three different values: Medicine Q will work for
none of the patients, one of the patients, or both patients. So the variable X is
equal to 0, 1, or 2.
148
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T04_RG.indd 148
9/17/11 3:43 AM
When X = 0, it means that Medicine Q failed to work for both patients.
Since P (success) = 0.7, P (failure) = 0.3.
REMEMBER
If A and B are independent events,
then P(A and B) = P(A) · P(B).
P(X = 0) = P (failure) · P (failure)
= 0.3 · 0.3
= 0.09
When X = 1, it means that the medicine worked for one of the patients,
but not both.
P(X = 1) = P (success) · P (failure) + P (failure) · P (success)
= 0.7 · 0.3 + 0.3 · 0.7
= 0.21 + 0.21
= 0.42
When X = 2, it means that the medicine worked for both patients.
P(X = 0) = P (success) · P (success)
= 0.7 · 0.7
= 0.49
We can now use the probabilities that we calculated for each outcome to
create a binomial distribution table and a binomial distribution histogram.
Binomial Distribution Table
X
0
1
2
P
0.09
0.42
0.49
Binomial Distribution Histogram
P
0.55
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
X
Interpreting Binomial Distributions
The doctor would like to be successful with at least one of the two patients.
In the language of probability, the question becomes what is P (X ≥ 1)?
The answer to this question is found by adding the appropriate probabilities
in the table.
P(X ≥ 1) = P (X = 1) + P (X = 2)
= 0.42 + 0.49
= 0.91
There is a 91% probability that Medicine Q will be successful with at least
one of the patients.
BINOMIAL DISTRIBUTIONS
149
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HS_PS_S1_04_T04_RG.indd 149
9/17/11 3:43 AM
Problem Set
Determine whether the variable X has a binomial distribution. If it has
a binomial distribution, then give P (success) and the number of trials.
If X does not have a binomial distribution, then explain why.
1.
When a coin is tossed, X is the number of tosses
it takes to get tails.
2.
Over the next 4 days, X is the number of days it
will rain.
3.
When a coin is tossed 4 times, X is the number
of tails.
4.
When a 6-sided number cube is rolled 3 times,
X is the number of 2s rolled.
5.
A bag contains a red, a yellow, and a blue marble.
When a marble is selected 2 times in a row
without replacement, X is the number of yellow
marbles.
6.
A bag contains a red, a yellow, and a blue marble.
When a marble is selected from the bag 3 times
in a row with replacement, X is the number of
yellow marbles.
Match the probability histogram with the description of the binomial
distribution. Assume that there are 2 trials.
A.
P (success) = 0.8
C.
P (success) = 0.2
B.
P (success) = 0.7
D.
P (success) = 0.3
7.
9.
0.7
0.5
0.6
0.4
0.5
0.3
0.4
P
P
0.6
0.2
0.3
0.1
0.2
0
0.1
0
8.
1
X
2
0
10.
1
X
2
0
1
X
2
0.7
0.5
0.6
0.4
0.5
0.3
0.4
P
P
0.6
0
0.2
0.3
0.1
0.2
0.1
0
0
150
CHAPTER 4
1
X
2
0
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T04_PS.indd 150
9/17/11 4:01 AM
Suppose the dial on the spinner is spun 2 times in a row. Create a
probability histogram for the variable X.
A
B
C
11.
X is the number of times the dial lands on
section A.
13.
X is the number of times the dial lands on
section A or C.
12.
X is the number of times the dial lands on
section C.
14.
X is the number of times the dial lands on
section B or A.
Create a probability distribution table for the variable X. Then use the
table to find the indicated probability.
15.
When a coin is tossed twice, X is the number of
tails. Find P (X < 2).
16.
A basketball player’s probability of making a
9
free throw is ___
10 . When the player shoots 2 free
throws in a row, X is the number of free throws
made. Find P (X = 2).
17.
When a coin is tossed 3 times, X is the number of
tails. Find P (X > 1).
18.
A basketball player’s probability of making a
9
free throw is ___
10 . When the player shoots 3 free
throws in a row, X is the number of free throws
made. Find P (X = 2).
19.
A basketball player’s probability of making a
free throw is 0.6. When the player shoots 2 free
throws in a row, X is the number of free throws
made. Find P (X = 1).
20.
A basketball player’s probability of making a
free throw is 0.6. When the player shoots 3 free
throws in a row, X is the number of free throws
made. Find P (1 ≤ X < 3).
Use the spinner to solve.
21.
D
A
C
B
Challenge Suppose the dial on the spinner is spun 3 times in a
row. Find the probability that the dial will land 2 or more times
on section B.
BINOMIAL DISTRIBUTIONS
151
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HS_PS_S1_04_T04_PS.indd 151
9/17/11 4:01 AM
Continuous Random
Variables
Areas within probability distributions can be used
to compute probability.
Exploring Continuous Random Variables
An incubator is being used to hatch chickens. The incubator temperatures X,
in degrees Fahrenheit, are distributed according to this probability distribution.
When all values of a random variable X
are equally likely to occur, the resulting
probability distribution is called a
uniform probability distribution.
For the incubator, all of the values
of X, from 100°F to 104°F are
equally likely to occur. Because any
temperature between these values,
such as 101.5°F or 103.9°F, may occur,
X, in this case, is considered to be a
continuous random variable.
P
0.5
Q&A
Uniform Probability
Distribution
Q
How is a discrete random
variable different from a
continuous random variable?
A
A discrete random variable
has a limited number of
values, while a continuous
random variable has an
infinite number of values.
0.4
0.3
0.2
0.1
X
0
99 100 101 102 103 104 105
CONTINUOUS RANDOM VARIABLE
A variable X is a continuous random variable if it can take on any value
in the possible range of the random variable.
Using Area to Determine Probability
When X is a continuous random variable, the resulting probability
distribution is a geometric shape with a total area of 1. The probability that
a ≤ X ≤ b will be the area of the shape bordered by the values a and b.
For instance, suppose the temperature of the incubator is taken at 10 a.m.
What is the probability that the temperature is between 101°F and 103°F?
152
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RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T05_RG.indd 152
9/17/11 4:06 AM
To answer this question, you can
compute the area of the rectangular
region where 101 ≤ X ≤ 103, which is
shown here.
The area of the rectangular region is
the product of the base and the height:
2 · 0.25 = 0.5. There is a 50%
probability that the temperature will be
between 101°F and 103°F degrees at
10 a.m.
P
REMEMBER
0.5
The area of the graph of a
probability distribution is
equal to 1.
Rectangular
region
0.4
0.3
0.2
0.1
X
0
99 100 101 102 103 104 105
Using Continuous Nonuniform Probability Distributions
Here is another continuous probability distribution that is not uniform and is
in the shape of a triangle.
P
Nonuniform Probability Distribution
0.6
0.4
0.2
0
X
1
2
3
4
5
6
Suppose we want to know P(1 ≤ X ≤ 4), the probability that X is between
1 and 4.
P
0.6
0.4
0.2
Trapezoidal
region
0
X
1
2
3
4
5
6
The region bounded by X = 1 and X = 4 is a trapezoid with base lengths 0.1
and 0.4. The height of the trapezoid is 3.
1
Area = __
2 h ( base1 + base2)
1
= __
2 · 3 (0.1 + 0.4)
= 0.75
So the probability that X is between 3 and 4 is 0.75, or 75%.
CONTINUOUS RANDOM VARIABLES
153
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HS_PS_S1_04_T05_RG.indd 153
9/17/11 4:06 AM
Problem Set
Use the uniform probability distribution to solve.
P
0
1
2
3
4
X
1.
What is the area of the distribution?
7.
Determine P (1.5 ≤ X ≤ 3).
2.
What is the height of the distribution?
8.
Determine P (1.5 < X ≤ 3).
3.
What is the area of the part of the region where
X ≤ 3?
9.
If P (X < a) = 0.5, what is the value of a?
4.
What is the area of the part of the region where
1.5 ≤ X ≤ 3?
10.
If P (X ≤ a) = 0.5, what is the value of a?
11.
2 , what is the value of a?
If P (X < a) = __
3
5.
Determine P (X ≤ 3).
6.
Determine P (X < 3).
12.
1 , what is the value of a?
If P (X ≤ a) = __
6
The weights X of an animal are distributed according to the probability
distribution shown.
X
147
148
149
150
151
152
153
13.
Suppose the animal is weighed at 11 a.m. What is probability that
the animal’s weight will be less than 150 pounds?
14.
Suppose the animal is weighed at 2 p.m. What is probability that the
animal’s weight will be less than 151 pounds?
15.
Suppose the animal is weighed in the morning. What is probability
1
that the animal’s weight will be within __
2 pound of 150 pounds?
16.
Suppose the animal is weighed in the afternoon. What is probability
3
that the animal’s weight will be within __
4 pound of 149 pounds?
154
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RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T05_PS.indd 154
9/17/11 4:17 AM
Use the probability distribution to solve.
P
0.3
0.2
0.1
0
1
2
3
4
5
6
7
8
X
17.
What is the area of the distribution?
23.
Determine P (1.5 < X ≤ 6.5).
18.
Determine P (X < 3).
24.
Determine P (X ≤ 8).
19.
Determine P (X ≤ 3).
25.
If P (X > a) = 0.5, what is the value of a?
20.
Determine P (X < 3 or X > 5).
26.
If P (X ≤ a) = 0.5, what is the value of a?
21.
Determine P (X ≤ 3 or X > 6).
27.
If P (X ≤ a) = 0.7, what is the value of a?
22.
Determine P (3 < X < 4).
28.
If P (X < a) = 0.7, what is the value of a?
A random number generator generates real numbers from 0 to 3,
according to this probability distribution.
29.
30.
What is the probability that a number between
1 and 2 is generated?
What is the probability that a number between
2 and 3 is generated?
P
0.6
0.5
31.
What is the probability that a number less than
2 is generated?
0.4
32.
What is the probability that a number between
0.5 and 2.5 is generated?
0.3
33.
What is the probability that a number less than
1 or greater than 1.5 is generated?
0.2
34.
Challenge Suppose the probability is 0.4 that
0.1
a number less than a is generated. What is the
value of a?
0
1
2
3
4
X
CONTINUOUS RANDOM VARIABLES
155
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HS_PS_S1_04_T05_PS.indd 155
9/17/11 4:17 AM
The Normal Distribution
The bell-shaped curve is often used for data analysis
and probability.
Using the Normal Distribution
When a data set has a normal distribution, a histogram that represents the
data set is symmetric and has a bell shape.
The Normal Distribution
FACT
In a normal distribution, the
mean, median, and mode all have
approximately the same value.
Mean
When a curve is drawn through the tops of the bars, the curve is in the shape
of a bell.
Normal distributions have important properties that help us study data sets
that are otherwise too large to study.
For example, the mean is always at the center of the distribution, so about
half of the data set is above the mean and half of the data set is below
the mean.
The standard deviation of a normal distribution also gives us useful
information about the data set.
PROPERTIES OF A NORMAL DISTRIBUTION
If a data set has a normal distribution, then about
• 68% of the data is within one standard deviation of the mean.
• 95% of the data is within two standard deviations of the mean.
• 99.7% of the data is within three standard deviations of the mean.
156
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T06_RG.indd 156
9/17/11 4:22 AM
The properties involving standard deviation are illustrated in more detail
in this diagram.
The Normal Distribution
99.7%
95%
REMEMBER
The Greek letters μ and σ are
used to represent the mean and
standard deviation of a data set,
respectively.
68%
2.35% 13.5%
34%
13.5% 2.35%
34%
μ - 3σ μ - 2σ μ - 1σ
μ
μ + 1σ μ + 2σ μ + 3σ
For example, we know that about 68% of the data is within one standard
deviation of the mean. This means that about 34% is one standard deviation
above the mean, and 34% of the data is one standard deviation below
the mean.
Using Normal Distribution Properties
A pediatrician read an article that states that the length of the flu incubation
period is distributed normally with a mean length of 7 days and a standard
deviation of 2 days. Afterward, he wonders what the probability is of the flu
incubating for at least 9 days.
Assuming that flu incubation periods are normally distributed, a normal
curve can be drawn to represent this situation.
REMEMBER
In a normal distribution, about
50% of the data is above the
mean, and about 50% of the data
is below the mean.
Flu Incubation Period
50%
3
5
34%
7
Number of days
9
11
We know that 9 days is one standard deviation above the mean of 7 days.
So, 34% of the data is between 7 days and 9 days. Also, 50% of the data is
below the mean.
This means that 50% + 34% = 84% of flu cases have an incubation period
of at most 9 days. By subtracting 84% from 100%, we can obtain the
percent of cases that have an incubation period of at least 9 days.
Flu Incubation Period
16%
3
5
7
Number of days
9
11
Therefore, the probability is about 16% that a flu case will have an
incubation period of at least 9 days.
THE NORMAL DISTRIBUTION
157
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HS_PS_S1_04_T06_RG.indd 157
9/17/11 4:22 AM
Problem Set
Suppose normal human body temperatures are normally distributed
with a mean of 37°C and a standard deviation of 0.2°C.
1.
What temperature is one standard deviation below the mean
temperature?
2.
What temperature is three standard deviations above the mean?
3.
What percent of humans have a temperature above 37°C?
4.
What percent of humans have a temperature below 37.2°C?
5.
What percent of humans have a temperature below 36.8°C?
6.
What percent of humans have a temperature between 36.8°C
and 37.2°C?
7.
What percent of humans have a temperature between 36.6°C
and 37.4°C?
8.
What percent of humans have a temperature between 37°C
and 37.4°C?
Suppose that the weights of 5400 registered female Labrador retrievers
in the United States are distributed normally with a mean of 62.5 pounds
and a standard deviation of 2.5 lb.
9.
What percent of the Labrador retrievers weigh less than 62.5 pounds?
10.
What percent of the Labrador retrievers weigh less than 65 pounds?
11.
What percent of the Labrador retrievers weigh between 60 pounds
and 65 pounds?
12.
What percent of the Labrador retrievers weigh between 60 lb and
62.5 pounds?
13.
Approximately how many of the Labrador retrievers weigh less
than 62.5 pounds?
14.
Approximately how many of the Labrador retrievers weigh less
than 65 pounds?
15.
Approximately how many of the Labrador retrievers weigh between
60 pounds and 65 pounds?
16.
Approximately how many of the Labrador retrievers weigh between
60 pounds and 62.5 pounds?
158
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T06_PS.indd 158
9/17/11 5:48 AM
Suppose that the number of petals on 2400 fully developed roses
is normally distributed with a mean of 32 petals and a standard
deviation of 2 petals.
17.
Approximately how many flowers have fewer than 32 petals?
18.
Approximately how many flowers have between 30 and 34 petals?
19.
Approximately how many flowers have between 28 and 32 petals?
20.
Approximately how many flowers have between 26 and 36 petals?
21.
Approximately how many flowers have 30 or more petals?
22.
Approximately how many flowers have fewer than 34 petals?
Suppose a factory produces 50,000 packages of mints each week.
Assume that the number of mints in a package is normally distributed
with a mean of 120 mints and a standard deviation of 3 mints.
23.
Approximately how many packages each week have between
120 and 129 mints?
24.
Approximately how many packages each week have between
111 and 117 mints?
25.
Approximately how many packages each week have more than
123 mints?
26.
Approximately how many packages each week have fewer than
126 mints?
Suppose a manufacturer makes nails that are normally distributed with
a length of 150 millimeters and a standard deviation of 1 millimeter.
27.
Challenge In a certain batch of nails, 3900 nails are between
28.
Challenge In a certain batch of nails, 25 nails are longer than
149 millimeters and 151 millimeters long. What is the approximate
number of nails in the entire batch? Round your answer to the
nearest whole number.
153 millimeters. What is the approximate number of nails in the
entire batch? Round your answer to the nearest whole number.
THE NORMAL DISTRIBUTION
159
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HS_PS_S1_04_T06_PS.indd 159
9/17/11 5:48 AM
Standardizing Data
A standard score tells us how far a data value is from
the mean.
Converting Raw Scores into z-Scores
Antibiotics are strong medicines that are prescribed by a doctor and often
used to treat infections. Once a patient begins taking an antibiotic, the
length of time that it takes to cure the infection can vary.
Suppose that the time it takes a certain antibiotic to cure an infection is
known to be normally distributed with a mean of 5.2 days and a standard
deviation of 1.1 days.
To get an idea of how effective the new antibiotic is with certain patients,
actual results can be converted into z-scores. A z-score tells us the number
of standard deviations that a data value is from the mean of the data set.
When a data value (called a raw score) is greater than the mean, its
corresponding z-score will be positive. When a data value is less than the
mean, its corresponding z-score will be negative.
REMEMBER
z-scores are sometimes called
standard scores.
CONVERTING RAW SCORES INTO z-SCORES
A z-score is the number of standard deviations that a data value is from
the mean. Every data value in a data set has a corresponding z-score and
is obtained by the formula
x−μ
z = _____
σ .
In this formula, z is the z-score, x is the raw data value, μ is the mean of
the data set, and σ is the standard deviation.
Suppose the antibiotic cures an infection in a patient in 3.8 days. What is
the corresponding z-score?
Here, x = 3.8, μ = 5.2, and σ = 1.1. These values can be substituted into
the formula to find the corresponding z-score.
3.8 − 5.2
z = ________
1.1
≈ −1.3
A z-score of −1.3 means that the antibiotic cures the infection in a time that
is 1.3 standard deviations below the mean.
160
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T07_RG.indd 160
9/17/11 4:31 AM
Converting z-Scores into Raw Scores
Suppose a patient takes the antibiotic and is cured of the infection according
to a z-score of 2.3. From what we know about z-scores, this means that the
patient was cured in an amount of time that is about 2.3 standard deviations
greater than the mean cure time.
But suppose we want to know how many days it took the antibiotic to cure
the infection in this patient. In other words, what is the raw score if the
z-score is 2.3?
The z-score formula can be rearranged to create a formula for converting
z-scores back into raw scores.
REMEMBER
A raw score refers to the original
data value.
CONVERTING z-SCORES INTO RAW SCORES
A z-score can be converted back into its original value, called a raw score,
using the formula
x = z ∙ σ + μ.
In this formula, x is the raw score, z is the z-score, μ is the mean of the
data set, and σ is the standard deviation.
Now, we can convert the z-score of 2.3 back into a raw score. Using z = 2.3,
μ = 5.2, and σ = 1.1, we can find the value of the raw score, x.
x = 2.3 · (1.1) + 5.2
≈ 7.7
For this patient, the antibiotic cured the infection in about 8 days—more than
2 days longer than the average cure time for this antibiotic.
STANDARDIZING DATA
161
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HS_PS_S1_04_T07_RG.indd 161
9/17/11 4:31 AM
Problem Set
Suppose a normal distribution has a mean of 20 and a standard
deviation of 4.
1.
A value of 24 is how many standard deviations away from the mean?
2.
A value of 26 is how many standard deviations away from the mean?
3.
What is the z-score of 22?
4.
What is the z-score of 13?
5.
What is the z-score of 30?
6.
What is the z-score of 0?
7.
What is the z-score of a value that is 1.43 standard deviations greater
than the mean?
8.
What is the z-score of a value that is 0.52 standard deviations less
than the mean?
Suppose the ages of cars driven by employees at a company are
normally distributed with a mean of 8 years and a standard deviation
of 3.2 years.
9.
What is the z-score of a car that is 12 years old?
10.
What is the z-score of a car that is 6 years old?
11.
What is the z-score of a car that is 9.1 years old?
12.
DeSean, a new employee, drives a brand new car. What is the
z-score of the car?
Suppose a manufacturer makes disposable peppercorn grinders. The
number of peppercorns in the grinders are normally distributed with a
mean of 322 peppercorns and a standard deviation of 5.3 peppercorns.
13.
Suppose a peppercorn grinder contains 315 peppercorns. What is the
z-score for this grinder?
14.
Suppose a peppercorn grinder contains 323 peppercorns. What is the
z-score for this grinder?
15.
Suppose the manufacturer will only sell peppercorn grinders with a
z-score between −2 and 2. What are the least and most peppercorns
a grinder can contain?
16.
Suppose the manufacturer will only sell peppercorn grinders with a
z-score between −0.9 and 0.9. What are the least and most
peppercorns a grinder can contain?
162
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T07_PS.indd 162
9/17/11 4:31 AM
Solve by completing the table. Round your answer to the
nearest hundredth.
17.
The healing time for a broken ankle is normally distributed with a
mean of 42 days and a standard deviation of 9.25 days.
Name
Ayla
Healing Time
35 days
Miranda
1.4
Stacy
18.
z-score
–2.2
With new technology, the healing time for a broken ankle now takes
a mean of 36 days with a standard deviation of 8.67 days.
Name
Alfredo
Josh
Sergio
Healing Time
z-score
35 days
1.4
–2.2
Suppose the velocities of golf swings for amateur golfers are normally
distributed with a mean of 95 miles per hour and a standard deviation
of 3.9 miles per hour.
19.
What is the difference in velocities between a golfer whose z-score
is 0 and another golfer whose z-score is 1?
20.
What is the difference in velocities between a golfer whose z-score
is 0 and another golfer whose z-score is −1?
21.
What is the difference in velocities between a golfer whose z-score
is −0.5 and another golfer whose z-score is 0.5?
22.
What is the difference in velocities between a golfer whose z-score
is −1 and another golfer whose z-score is −2?
23.
What is the difference in velocities between a golfer with a z-score
of a and another golfer whose z-score is a + 1?
Suppose a manufacturer makes nails that are normally distributed
with a mean length of x millimeters and a standard deviation of
y millimeters.
24.
Challenge In a certain batch of nails, an 83-millimeter nail has
a z-score of 1 and an 84.5-millimeter nail has a z-score of 2.
Determine x and y.
STANDARDIZING DATA
163
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HS_PS_S1_04_T07_PS.indd 163
9/24/11 2:08 AM
Comparing Scores
Standard scores can be used to compare
normal distributions.
Comparing z-Scores from Two Distributions
Black tea has been studied for its medicinal properties and for the amount of
caffeine it contains (which can vary depending on the type of tea used and
the amount of time brewed).
The amount of caffeine in single servings of tea is normally distributed.
For a particular comparison, single servings of two different Ceylon tea
mixtures are compared.
Mixture A
BY THE WAY
Ceylon tea is a type of black tea
grown in Sri Lanka.
Mixture B
Mean: μ = 55 mg
Mean: μ = 52 mg
Standard Deviation: σ = 4.1 mg
Standard Deviation: σ = 3.3 mg
Suppose you are given a serving of Ceylon tea which contains 50 milligrams
of caffeine. Which mixture is the tea more likely to come from?
To determine whether the serving with 50 milligrams of caffeine is more
likely to come from Mixture A or Mixture B, you can determine the z-score
for each mixture.
Mixture A
Mixture B
50 − 55
z = _______
4.1
50 − 52
z = _______
3.3
≈ −1.22
≈ −0.61
For a normal distribution, any value that is closer to the mean will be a
more likely outcome than any value farther away from the mean. Since
−0.61 is closer to 0 than −1.22, Mixture B is more likely to produce a
serving of Ceylon tea with 50 milligrams of caffeine.
Paulo is doing his own research about the caffeine content of these two
Ceylon tea mixtures. The brew he makes with Mixture A has 57 milligrams
of caffeine in a single serving. The brew he makes with Mixture B has
49 milligrams of caffeine in a single serving. Which result is more unusual?
164
CHAPTER 4
REMEMBER
x−μ
z = _____
σ
THINK ABOUT IT
The z-score of the mean is 0.00
because it is zero standard
deviations away from the mean.
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T08_RG.indd 164
9/17/11 4:36 AM
To determine whether the serving with 57 milligrams of caffeine from
Mixture A is more unusual than the serving with 49 milligrams of caffeine
from Mixture B, you can determine the z-score for each mixture.
Mixture A
Mixture B
57 − 55
z = _______
4.1
49 − 52
z = _______
3.3
≈ 0.49
≈ −0.91
The z-scores for each mixture have different signs (one positive and one
negative). However, we are only interested in knowing which z-score is
farthest from 0.
Since Mixture A’s z-score of 0.49 is closer to 0 than Mixture B’s z-score of
−0.91, the serving with 49 milligrams of caffeine from Mixture B is more
unusual than the serving with 57 milligrams of caffeine from Mixture A.
Interpreting a z-Score in Two Distributions
Two normally distributed data sets can also be compared for the same
z-score. For instance, in looking at the two mixtures of Ceylon tea, which
mixture has more caffeine for a z-score of −3.00?
We know that the z-score is −3.00, and we know the mean and standard
deviation for each mixture. We can use the appropriate formula to determine
the actual amount of caffiene in a serving from each mixture.
Mixture A
Mixture B
x = −3 · (4.1) + 55
x = −3 · (3.3) + 52
= 42.7
REMEMBER
x=z·σ+μ
= 42.1
At a distance of three standard deviations below the mean, the two mixtures
have about the same amounts of caffeine. However, Mixture A has more
caffeine compared to Mixture B.
COMPARING SCORES
165
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HS_PS_S1_04_T08_RG.indd 165
9/17/11 4:36 AM
Problem Set
The bar graph shows the z-score results of four students on
two different math tests. The students took Test 1, and then
a month later took Test 2.
Math Test z-Scores
1.75
1.5
1.25
1
z-Score
0.75
0.5
0.25
0
−0.25
−0.5
−0.75
Test 1
−1
Test 2
−1.25
Carla
Dave
Euan
Students
Felicia
1.
What are Carla’s z-scores for both tests?
4.
2.
How many standard deviations away from the
mean did Dave score on Test 2? Was the score
above the mean or below the mean? Explain.
Which student had the lowest score on Test 2?
Explain.
5.
Which student had the best overall performance
on both tests? Explain.
3.
Which student scored the farthest away from the
mean? On which test did this occur? Explain.
6.
Which student showed the most improvement
from Test 1 to Test 2? Explain.
Suppose the growth of Bermuda grass from germination to lawn care
length is normally distributed with a mean of 20 days with a standard
deviation of 3.3 days. The growth of buffalo grass from germination to
lawn care length is normally distributed with a mean of 21 days with a
standard deviation of 2.3 days.
7.
A gardener grew a lawn using Bermuda grass in
23 days. Determine the z-score.
8.
A gardener grew a lawn using buffalo grass in
15 days. Determine the z-score.
166
CHAPTER 4
9.
10.
On which lawn would a z-score of 3 give a
longer growing period? Explain.
On which lawn would a z-score of −2.5 give a
longer growing period? Explain.
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T08_PS.indd 166
9/17/11 4:39 AM
Suppose scores on a biology test are normally distributed with a
mean of 85 and a standard deviation of 4. Scores on a history test are
normally distributed with a mean of 85 and a standard deviation of 2.
11.
Ayla scored 90 on both tests. Determine her z-score for each test.
On which test did she have the higher z-score?
12.
Maya scored 80 on both tests. Determine her z-score for each test.
On which test did she have the higher z-score?
13.
Eric scored 89 on the biology test and 87 on the history test. On which
test did he have the better performance? Explain.
14.
Paulo scored 90 on the biology test and 89 on the history test. On which
test did he have the better performance? Explain.
15.
On which test would a z-score of 3 give a higher raw score? Explain.
16.
On which test would a z-score of −2.5 give a higher raw score? Explain.
17.
On which test would a z-score of 0.5 give a higher raw score? Explain.
18.
On which test would you expect a higher range of scores? Explain.
19.
On which test would a score of 87 be a better score? Explain.
20.
On which test would a score of 80 be a better score? Explain.
Suppose Company A produces packages of throat lozenges that are
normally distributed with a mean of 38.2 individual lozenges and a
standard deviation of 1.7 lozenges. Company B produces packages
of throat lozenges that are normally distributed with a mean of
36.9 individual lozenges and a standard deviation of 2.2 lozenges.
21.
Which company would be more likely to produce a package of 37 throat
lozenges? Explain using z-scores.
22.
Which company would be more likely to produce a package of 43 throat
lozenges? Explain using z-scores.
23.
Would it be more likely for Company A to produce a package of 35 throat
lozenges or Company B to produce a package of 40 throat lozenges?
Explain using z-scores.
24.
Challenge Suppose each company produces a package of x lozenges so
that the z-scores are identical. Find the value of x.
COMPARING SCORES
167
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HS_PS_S1_04_T08_PS.indd 167
9/17/11 4:39 AM
The Standard Normal Curve
The standard normal curve is used to analyze data.
Standard Normal Curve Properties
When all of the data values of a normal distribution are turned into z-scores,
the resulting distribution is called the standard normal curve.
Each value on the horizontal axis of the
standard normal curve is a z-score, and the
mean is always 0.
Standard Normal Curve
For instance, a data value with a z-score
of 0.54 would correspond to 0.54 on the
horizontal axis of the standard normal curve.
–3
–2
–1 0
1
z-scores
2
3
PROPERTIES OF THE STANDARD NORMAL CURVE
The standard normal curve is a probability distribution with
the following properties:
• The mean is 0.
• The standard deviation is 1.
• The area under the curve is 1.
Using the z-Distribution Table
Because the standard normal curve is a probability distribution, it can be
used to find probabilities. These probabilities correspond to areas under the
standard normal curve.
Suppose a data value has a z-score of 0.54. What percent of the data values
are less than this value?
REMEMBER
If a z-score is positive, then it is
above the mean. If a z-score is
negative, then it is below the
mean.
?
0 0.54
z-scores
168
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T09_RG.indd 168
9/17/11 4:44 AM
To answer this question, we can use the Table of z-Scores for Normal
Distribution (pages A-1 and A-2). This table gives the areas under the
standard normal curve that are less than a given z-score.
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
.00
0.5000
0.5398
0.5793
0.6179
0.6554
0.6915
0.7257
.01
0.5040
0.5438
0.5832
0.6217
0.6591
0.6950
0.7291
.02
0.5080
0.5478
0.5871
0.6255
0.6628
0.6985
0.7324
.03
0.5120
0.5517
0.5910
0.6293
0.6664
0.7019
0.7357
.04
0.5160
0.5557
0.5948
0.6331
0.6700
0.7054
0.7389
BY THE WAY
.05
0.5199
0.5596
0.5987
0.6368
0.6736
0.7088
0.7422
Using the z-distribution table, look for
the area that corresponds to the row and
column of a z-score of 0.54. We can see that
P (z < 0.54) = 0.7054, which is about 71%.
Technology, such as calculators
and computers, can also be used
to find areas under the standard
normal curve.
Area = 0.7054
This means that about 71% of the data values
are less than a value with a z-score of 0.54.
0 0.54
z-scores
Using the Standard Normal Curve to Solve Problems
Suppose scores on a standardized test for future doctors are normally
distributed with a mean of 10.5 and a standard deviation of 2.2. If 150 future
doctors take the test, how many are likely to score between 12 and 15?
The score 12 has a z-score of about 0.68, and the score 15 has a z-score of
about 2.05. To help us answer the question, we need to know the area that
corresponds to P (0.68 < z < 2.05).
REMEMBER
z-scores are obtained using the
following formula:
x−μ
z = _____
σ
The area under the standard normal curve less than z = 0.68 is 0.7517, and
the area less than z = 2.05 is 0.9798.
So P (0.68 < z < 2.05) = 0.9798 − 0.7517 = 0.2281.
0.2281
0 0.68
z-scores
2.05
Multiply the area by the number of doctors taking the exam.
0.2281 · 150 ≈ 34
So about 34 of the 150 future doctors taking the exam are likely to score
between 12 and 15.
THE STANDARD NORMAL CURVE
169
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HS_PS_S1_04_T09_RG.indd 169
9/17/11 4:44 AM
Problem Set
The expression represents the area under the standard normal curve
below a given value of z. Use the z-distribution table on pp. A-1 to A-2
or technology to find the indicated area, and include a sketch of the
standard normal curve with the appropriate area shaded.
1.
P (z < 0.00)
5.
P (z < −1.00)
2.
P (z < 1.00)
6.
P (z < −1.53)
3.
P (z < 1.57)
7.
P (z < −2.02)
4.
P (z < 2.87)
8.
P (z < −2.77)
The expression represents the area under the standard normal curve
above a given value of z. Use the z-distribution table on pp. A-1 to A-2
or technology to find the indicated area, and include a sketch of the
standard normal curve with the appropriate area shaded.
9.
P (z > 0.00)
13.
P (z > −1.00)
10.
P (z > 1.00)
14.
P (z > −1.53)
11.
P (z > 1.57)
15.
P (z > −2.05)
12.
P (z > 2.87)
16.
P (z > −2.87)
The expression represents the area under the standard normal curve
above a given value of z. Use the z-distribution table on pp. A-1 to A-2
or technology to find the indicated area, and include a sketch of the
standard normal curve with the appropriate area shaded.
17.
P (0.00 < z < 1.00)
21.
P (−1.33 < z < 1.59)
18.
P (0.00 < z < 2.00)
22.
P (−2.00 < z < 2.00)
19.
P (1.00 < z < 1.00)
23.
P (−1.61 < z < −0.09)
20.
P (1.06 < z < 2.63)
24.
P (−2.53 < z < −0.55)
170
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T09_PS.indd 170
9/24/11 2:09 AM
The scores on a standardized test are normally distributed with a mean
of 500 and a standard deviation of 100. Use the z-distribution table on
pp. A-1 to A-2 or technology to solve.
25.
Sofia scored 632 on the test. What percent of students scored
below Sofia?
26.
Jake scored 500 on the test. What percent of students scored
below Jake?
27.
Benita scored 432 on the test. What percent of students scored
below Benita?
28.
Kendall scored 380 on the test. What percent of students scored
below Kendall?
29.
Suppose 6000 students took the test. Approximately how many
students scored below 632?
30.
Suppose 6000 students took the test. Approximately how many
students scored above 490?
31.
Suppose 6000 students took the test. Approximately how many
students scored between 450 and 550?
32.
Suppose 6000 students took the test. Approximately how many
students scored between 500 and 600?
Lengths of newborn girls are normally distributed with a mean of
49.2 centimeters and a standard deviation of 1.8 centimeters. Use the
z-distribution table on pp. A-1 to A-2 or technology to solve.
33.
What percent of newborn girls will be 50 centimeters or longer?
34.
What percent of newborn girls will be 49 centimeters or longer?
35.
What percent of newborn girls will be 52 centimeters or longer?
36.
What percent of newborn girls will be 45 centimeters or shorter?
37.
Consider a group of 2000 newborn girls. How many girls will be
50.5 centimeters or longer?
38.
Consider a group of 2000 newborn girls. How many girls will be
51 centimeters or shorter?
39.
Consider a group of 2000 newborn girls. How many girls will be
between 49 centimeters and 51 centimeters?
40.
Consider a group of 2000 newborn girls. How many girls will be
between 44 centimeters and 45 centimeters?
41.
Challenge Consider a group of 2000 newborn girls. If 500 of the
newborn girls are between 47 centimeters and x centimeters long,
what is the value of x?
THE STANDARD NORMAL CURVE
171
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Finding Standard Scores
Areas under the standard normal curve can be used
to find standard scores and percentiles.
Using Area To Find a z-Score
While the straightforward way of reading the Table of z-Scores Normal
Distribution is to look up the z-score and then find the corresponding area
under the standard normal curve, we can also read the table in the other
direction.
REMEMBER
z-scores are sometimes called
standard scores.
Suppose that about 20% of the data in a data set is below a certain value.
What z-score corresponds to this value?
Standard Normal Curve
20%
z=?
z-scores
Since 20% = 0.2, we want to find an area in the body of the z-distribution
table that is closest to 0.2.
z
–1.4
–1.3
–1.2
–1.1
–1.0
–0.9
–0.8
–0.7
.00
0.0808
0.0968
0.1151
0.1357
0.1587
0.1841
0.2119
0.2420
.01
0.0793
0.0951
0.1131
0.1335
0.1562
0.1814
0.2090
0.2389
.02
0.0778
0.0934
0.1112
0.1314
0.1539
0.1788
0.2061
0.2358
.03
0.0764
0.0918
0.1093
0.1292
0.1515
0.1762
0.2033
0.2327
.04
0.0749
0.0901
0.1075
0.1271
0.1492
0.1736
0.2005
0.2296
.05
0.0735
0.0885
0.1056
0.1251
0.1469
0.1711
0.1977
0.2266
The area 0.2005 is closest to 20%. Use the row and column of the area to
locate the corresponding z-score. This corresponds to a z-score of −0.84.
172
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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Determining Area Above a z-Score
Suppose that 15% of the data in a data set is above a certain value.
What z-score corresponds to this value?
85%
15%
z-scores
z=?
Since the z-distribution table only gives the areas that are below a
particular z-score, we need to find the percent of data below the value:
100% − 15% = 85%. Since 85% = 0.85, we want to find an area in the
body of the z-distribution table that is closest to 0.85.
z
0.6
0.7
0.8
0.9
1.0
1.1
.00
0.7257
0.7580
0.7881
0.8159
0.8413
0.8643
.01
0.7291
0.7611
0.7910
0.8186
0.8438
0.8665
.02
0.7324
0.7642
0.7939
0.8212
0.8461
0.8686
.03
0.7357
0.7673
0.7967
0.8238
0.8485
0.8708
.04
0.7389
0.7704
0.7995
0.8264
0.8508
0.8729
.05
0.7422
0.7734
0.8023
0.8289
0.8531
0.8749
The area 0.8508 is closest to 85%. This corresponds to a z-score of 1.04.
Understanding Percentiles
A percentile rank is the percentage of data that falls below a particular
value. For example, if a student scores higher than 60% of the other
students who took an exam, then the student’s percentile rank is at the
60th percentile.
Scores on the math section of the Scholastic Aptitude Test (SAT) are
normally distributed with a mean of about 500 and a standard deviation of
about 100. According to this information, what score would be at the
80th percentile?
Remember
The nth percentile is the value
(or score) below which n percent
of the data may be found.
Using the z-distribution table, we see that the area 0.8023 is closest to 80%.
This corresponds to a z-score of 0.85. We can use the appropriate formula to
find the corresponding score.
x = z · σ + µ
= 0.85 · 100 + 500
= 585
So a score of 585 on the math section of the SAT is at the 80th percentile.
FINDING STANDARD SCORES 173
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Problem Set
The number represents the area under the standard normal curve
below a particular z-score. Find the z-score, and include a sketch of the
standard normal curve with the appropriate area shaded.
1.
0.2358
5.
0.9115
2.
0.9967
6.
0.8869
3.
0.5000
7.
0.1003
4.
0.1170
8.
0.2389
Suppose a set of data is normally distributed. Use the z-distribution
table on pp. A-1 to A-2 or technology to solve.
9.
Below which z-score do approximately half of the data lie?
10.
Above which z-score do approximately half of the data lie?
11.
Below which z-score do approximately 23.58% of the data lie?
12.
Above which z-score do approximately 23.58% of the data lie?
13.
Below which z-score do approximately 88.3% of the data lie?
14.
Above which z-score do approximately 88.3% of the data lie?
15.
3
Above which z-score do approximately __
4 of the data lie?
16.
3
Below which z-score do approximately __
4 of the data lie?
Suppose scores on a standardized test are normally distributed with
a mean of 500 and a standard deviation of 100. Use the z-distribution
table on pp. A-1 to A-2 or technology to solve.
17.
Approximately 50% of the scores are above
which z-score? Which test score corresponds
to the z-score?
20.
Approximately 10% of the scores are below
which z-score? Which test score corresponds
to the z-score?
18.
Approximately 50% of the scores are below
which z-score? Which test score corresponds
to the z-score?
21.
Approximately 80% of the scores are above
which z-score? Which test score corresponds
to the z-score?
19.
Approximately 10% of the scores are above
which z-score? Which test score corresponds
to the z-score?
22.
Approximately 80% of the scores are below
which z-score? Which test score corresponds
to the z-score?
174
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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HS_PS_S1_04_T10_PS.indd 174
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Suppose ages of cars driven by company employees are normally
distributed with a mean of 8 years and a standard deviation of
3.2 years. Use the z-distribution table on pp. A-1 to A-2 or
technology to solve.
23.
Approximately 15% of cars driven by company employees are older
than what age?
24.
Approximately 75% of cars driven by company employees are older
than what age?
25.
In a sample of 8000 company employees who drive cars, 6000 of
them drive cars younger than x years old. Determine x.
26.
In a sample of 10,000 company employees who drive cars, 1350 of
them drive cars younger than y years old. Determine y.
Assume that lengths of newborn girls are normally distributed with a
mean of 49.2 centimeters and a standard deviation of 1.8 centimeters.
27.
Find the length of a newborn girl whose length is at the 50th percentile.
28.
Find the length of a newborn girl whose length is at the 90th percentile.
29.
Find the length of a newborn girl whose length is at the 45th percentile.
30.
What is the percentile rank of a baby girl whose length is 51 centimeters?
31.
What is the percentile rank of a baby girl whose length is 53 centimeters?
32.
What is the percentile rank of a baby girl whose length is 47 centimeters?
FINDING STANDARD SCORES
175
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HS_PS_S1_04_T10_PS.indd 175
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Chapter 4 Wrap-Up
Probability Distributions
Though there are slightly more boys than girls under the age of 15 in
the United States, it is often estimated that there is a 50% chance that a
newborn baby will be a boy (or girl as the case may be).
In general, there should be an equal probability of a newborn being either a
boy or a girl. To test this, a binomial distribution can be created for a family
with 3 children.
This is a binomial distribution with P(success) = 0.5 and 3 trials. The
probability distribution is given in the table and the histogram.
X
0
1
2
3
P
0.125
0.375
0.375
0.125
Number of Boys in Family
(Predicted)
40
35
Probability (%)
30
25
20
15
10
5
0
0
1
2
Number of boys
3
A few conclusions can be drawn from this distribution. In a family of
3 children,
• There is a 37.5% chance that the family will have 1 boy.
• There is a 75% chance that the family will have boys and girls.
• There is a 25% chance that the family will have all boys or all girls.
176
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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Normal Distributions
In general, the length of a full-term newborn baby is distributed normally
with a mean of 20.1 inches with a standard deviation of 0.4 inches. According
to the properties of a normal distribution, this means that approximately 68%
of newborn babies are between 19.7 inches and 20.5 inches; 95% of newborn
babies are between 19.3 inches and 20.9 inches; and 99.7% of newborn
babies are between 18.9 inches and 21.3 inches.
What is the 80th percentile of a newborn baby’s length? According to the
standard normal distribution, the value in the z-distribution table closest
to 0.8 is when the z-score is z = 0.84.
To convert the z-score into the actual length, multiply 0.84 by the standard
deviation (0.4), then add the mean (20.1): 0.84 · (0.4) + 20.1 = 20.44.
This means that 80% of newborn babies will be no more than
20.44 inches long.
In Summary
Probability distributions give probabilities over a whole set of outcomes.
A binomial distribution explains repeated trials of the same event when the
probability of success is the same for each trial. A normal distribution has
properties that can be used to analyze data sets.
CHAPTER 4 WRAP-UP
177
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Practice Problems
Random Variables and Distributions
The frequency table shows the summer employment hourly wages
(to the nearest dollar) for a group of high school students.
Hourly Wage
$7
$8
$9
$10
$11
$12
$17
Number of High School Students
6
4
3
4
1
1
1
1.
Explain why the table is not a probability distribution table.
2.
If X is the hourly wage, complete the probability distribution table.
Round probabilities to the nearest hundredth.
X
7
8
9
10
11
12
17
P
3.
If a high school student is selected at random, determine P(X < 10).
4.
Based on these data, what would be the expected value of a high
school student’s summer hourly wage?
The Personal Computers data set on p. A-9 shows the number of
personal computers owned per 100 inhabitants of South American
countries from 1998 to 2005.
5.
Create a probability distribution table for 2004. Round probabilities
to the nearest thousandth.
6.
Suppose a personal computer is in use in South America in 2004.
What is the probability, to the nearest thousandth, that the computer
is being used in Chile?
7.
Suppose there are 280,000 personal computers of a certain brand in
use in South America in 2004. How many would you expect to be
used in Peru?
Determine if X has a binomial distribution. If it does, create a binomial
distribution histogram. Otherwise, explain why X does not have a
binomial distribution.
8.
When a coin is tossed twice, X is the number of heads.
9.
A tennis player’s probability of winning a match is 0.6. X is the
number of wins for the next 3 matches.
10.
178
Over the next 5 days, X is the number of days it will snow.
CHAPTER 4
RANDOM VARIABLES AND DISTRIBUTIONS
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Chapter 4 Wrap-Up
The uniform probability distribution shows temperature X of an oven
set at 325°F at random times.
318
320
11.
What is the area of the distribution?
12.
What is the height of the distribution?
13.
Determine P (X < 323).
322
324
326
328
Temperature (°F)
330
332
X
14.
Suppose the oven temperature is taken at
6 p.m. What is probability that the temperature
will be within 1.5°F of 325°F?
15.
If P (X > a) = 0.7, what is the value of a?
Use the probability distribution to find the probability indicated.
16.
What is the area of the distribution?
17.
Determine P (X < 3).
18.
Determine P (X < 2 or X > 4).
19.
Determine P (3 < X ≤ 6).
20.
If P (X < a) = 0.5, what is the value of a?
21.
If P (X < a) = 0.25, what is the value of a?
P
0.3
0.25
0.2
0.15
0.1
0.05
0
X
1 2 3 4 5 6 7
Use the z-distribution table on pp. A-1 to A-2 or technology to solve.
The number of seeds in a certain type of strawberry are distributed
normally with a mean of 200 seeds and a standard deviation of 9 seeds.
22.
What percent of strawberries have fewer than
200 seeds?
25.
If X is the number of seeds on a strawberry,
what is P (185 < X < 205)?
23.
What percent of strawberries have more than
191 seeds?
26.
How many seeds would a strawberry contain if it
were at the 20th percentile?
24.
If a strawberry has a z-score of 1.11, about how
many seeds does it contain?
27.
How many seeds would a strawberry contain if it
were at the 90th percentile?
Use the z-distribution table on pp. A-1 to A-2 or technology to solve.
Scores on a college entrance test of verbal skills are normally
distributed with a mean of 78 and a standard deviation of 5. At the
same college, scores on the entrance test of mathematical skills are
normally distributed with a mean of 84 and a standard deviation of 2.
28.
Henry had a z-score of −0.1 on the verbal skills test and a z-score of 1.8
on the mathematical skills test. What raw score did he get on each test?
29.
Millie scored an 83 on the verbal skills test and an 86 on the
mathematical skills test. Explain why Millie did equally well on
both tests.
30.
On which test is a score of 80 more likely? Explain.
CHAPTER 4 WRAP-UP
179
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